CODEWITHSUNDEEP
phpjqueryajax
user1385745
user1385745

Reputation: 33

return php variable to jquery ajax

I have an ajax function in jquery calling a php file to perform some operation on my database, but the result may vary. I want to output a different message whether it succeeded or not

i have this : 

echo '<button id="remove_dir" onclick="removed('.$dir_id.')">remove directory</button>';

<script type="text/javascript">
    function removed(did){
        $.ajax({ 
            type: "POST", 
            url: "rmdir.php", 
            data: {dir_id: did},
            success: function(rmd){ 
                if(rmd==0)
                    alert("deleted");
                else
                    alert("not empty");
                window.location.reload(true);
            } 
        }); 
    }
</script>

and this 

   <?php
require('bdd_connect.php');
require('functions/file_operation.php');
if(isset($_POST['dir_id'])){
    $rmd=remove_dir($_POST['dir_id'],$bdd);
}
?>

my question is, how to return $rmd so in the $.ajax, i can alert the correct message ?

thank you for your answers

Upvotes: 3

Views: 17380

Answers (5)

Sinetheta
Sinetheta

Reputation: 9449

You need your php file to send something back, then you need the ajax call on the original page to behave based on the response.

php:

if(isset($_POST['dir_id'])){
    $rmd=remove_dir($dir_id,$bdd);  
    echo "{'rmd':$rmd}";
}

which will output one of two things: {"rmd": 0} or {"rmd": 1}

We can simulate this return on jsBin

Then use jquery to get the value and do something based on the response in our callback:

$.ajax({ 
    type: "POST",
    dataType: 'json',
    url: "http://jsbin.com/iwokag/3",
    success: function(data){ 
        alert('rmd = ' + data.rmd)
    } 
});

View the code, then watch it run. Only I didn't send any data here, my example page always returns the same response.

Upvotes: 1

VVLeon
VVLeon

Reputation: 399

Try echo the $rmd out in the php code, as an return to the ajax.

if(isset($_POST['dir_id'])){
    $rmd=remove_dir($dir_id,$bdd);  
    //if $rmd = 1 alert('directory not empty');
    //if $rmd = 0 alert('directory deleted');
    echo $rmd;
}

Your "rmd" in success: function(rmd) should receive the callabck.

Upvotes: 0

adeneo
adeneo

Reputation: 318372

PHP

<?php
   require('bdd_connect.php');
   require('functions/file_operation.php');
   if (isset($_POST['dir_id'])){
      $rmd=remove_dir($dir_id,$bdd); 
      echo $rmd;
   }
?>

JS

function removed(did){
    $.ajax({ 
        type: "POST", 
        url: "rmdir.php", 
        data: {dir_id: did}
    }).done(function(rmd) {
         if (rmd===0) {
            alert("deleted");
         }else{
            alert("not empty");
            window.location.reload(true);  
         }
    });
}

Upvotes: 5

Doug Molineux
Doug Molineux

Reputation: 12431

Just try echoing $rmd in your ajax file, and then watching the console (try console.log(rmd) in your ajax response block)

$.ajax({ 
        type: "POST", 
        url: "rmdir.php", 
        data: {dir_id: did},
        success: function(rmd){ 

            console.log(rmd);
        } 
    });

You can then act accordingly based on the response

Upvotes: 0

mgraph
mgraph

Reputation: 15338

i advice to use json or :

if(isset($_POST['dir_id'])){
    $rmd=remove_dir($dir_id,$bdd);  
    echo $rmd;
}

Upvotes: 3

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